Denote by ppp the greatest prime number not exceeding nnn. By Bertrand’s postulate there is a primeqqq with
p<q<2⁢ppq2pp<q<2p. Therefore we have n<2⁢pn2pn<2p. If HnsubscriptHnH_{n} were an integer, then the sum

had to be divisible by ppp. However its addend n!pnp\displaystyle\frac{n!}{p} is not divisible by ppp but all other addends are, whence the sum cannot be divisible by ppp. The contradictory situation means that HnsubscriptHnH_{n} is not integer when n>1n1n>1.

Theorem (Wolstenholme). If ppp is a prime number greater than 3, then the numerator of the harmonic number

where a1,a2,…,ap-2subscripta1subscripta2normal-…subscriptap2a_{1},\,a_{2},\,\ldots,\,a_{{p-2}} are integers. Because 1, 2,…,p-11 2normal-…p11,\,2,\,\ldots,\,p\!-\!1 form a set of all modulo ppp incongruent roots of the Fermat’s congruencexp-1≡1(modp)superscriptxp1annotated1pmodpx^{{p-1}}\equiv 1\;\;(\mathop{{\rm mod}}p), one may write the identical congruence

Since p∣ap-3=f′′⁢(0)2fragmentspnormal-∣subscriptap3superscriptf′′02p\mid a_{{p-3}}=\frac{f^{{\prime\prime}}(0)}{2}, one has p∣f′′(0)fragmentspnormal-∣superscriptf′′fragmentsnormal-(0normal-)p\mid f^{{\prime\prime}}(0). It then follows by (6) that p2∣f′(0)fragmentssuperscriptp2normal-∣superscriptfnormal-′fragmentsnormal-(0normal-)p^{2}\mid f^{{\prime}}(0). And since (5) divided by -(p-1)!p1-(p\!-\!1)! gives